This preview shows pages 1–3. Sign up to view the full content.
1
2.3. EBRT, IBRT in A,fA,fU.
We redo section 2.2 for the signature A,fA,fU, with the
same five BRT settings (SD,INF), (ELG
∩
SD,INF), (ELG,INF),
(EVSD,INF), (MF,INF).
After we treat these five BRT settings, we then treat the
five corresponding unary BRT settings (SD[1],INF), (ELG[1]
∩
SD[1],INF), (ELG[1],INF), (EVSD[1],INF), (MF[1],INF).
These are the same except that we restrict to the 1ary
functions only. There is quite a lot of difference between
the unary settings and the multivariate settings; this was
not the case in section 2.2, with just A,fA.
We begin with EBRT in A,fA,fU. The 8 A,fA,fU pre elementary
inclusions are as follows (see Definition 1.1.35).
A
∩
fA
∩
fU =
∅
.
A
∪
fA
∪
fU = U.
A
⊆
fA
∪
fU.
fA
⊆
A
∪
fU.
fU
⊆
A
∪
fA.
A
∩
fA
⊆
fU.
A
∩
fU
⊆
fA.
fA
∩
fU
⊆
A.
The 6 A,fA,fU elementary inclusions are as follows (see
Definition 1.1.36).
A
∩
fA =
∅
.
A
∪
fU = U.
A
⊆
fU.
fU
⊆
A
∪
fA.
A
∩
fU
⊆
fA.
fA
⊆
A.
We will use Theorem 2.2.1, and the Complementation Theorem
from section 1.3. In fact, we need the following
strengthening of the Complementation Theorem.
THEOREM 2.3.1. Let f
∈
SD and B
∈
INF. There exists A
∈
INF, A
⊆
B, such that A
∩
fA =
∅
and B
⊆
A
∪
fA. Moreover,
this is provable in RCA
0
.
Proof: Let f,B be as given. We inductively define A
⊆
B as
follows. Suppose the elements of A from 0,1,.
..,n1 have
been defined, n
≥
0. We put n in A if and only if n
∈
B and
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document2
n is not the value of f at arguments from A less than n.
Then A is as required, using f
∈
SD. QED
THEOREM 2.3.2. For all f
∈
EVSD there exists A
∈
INF such
that A
∩
fA =
∅
, A
∪
fN = N. Moreover, this is provable in
RCA
0
.
Proof: Let f
∈
EVSD be kary. Let n be such that x
≥
n
→
f(x)
≥
x. We define A inductively. First put [0,n]\fN in
A. For m > n, put m in A if and only if m = f(x) for no x
< m. Then [n+1,
∞
)
⊆
A
∪
fA. Also [0,n]
⊆
A
∪
fA. Hence A
∪
fN = N. Suppose m
∈
A
∩
fA. If m
≤
n then by construction,
m
∈
[0,n]\fN, contradicting m
∈
fA. Hence m > n. Let m =
f(x), x
∈
A
k
. If x
≥
m then f(x) > x
≥
m, which is a
contradiction. Hence x < m, and so m
∉
A by construction.
This contradicts m
∈
A. QED
THEROEM 2.3.3. Let k
≥
2. There exists kary f
∈
ELG
∩
SD
such that N\fN = {0}. There exists kary f
∈
ELG such that
fN = N.
Proof: For all n
≥
1, let f
n
:[2
n
,2
n+1
)
k
→
[2
n+1
,2
n+2
) be onto.
Let f be the union of the f
n
extended as follows. For x not
yet defined, set f(x) = 1 if x = 0; 2 if x = 1; 3 if
x = 2; 2x if x
≥
3. Then fN = N\{0} and f
∈
ELG
∩
SD.
Let g be the union of the f
n
extended as follows. For x not
yet defined, set f(x) = 0 if x = 0; 1 if x = 1; 2 if
x = 2; 3 if x = 3; 2x if x
≥
4. Then fN = N and f
∈
ELG. QED
SETTINGS: (SD,INF), (ELG
∩
SD,INF),
(ELG,INF), (EVSD,INF), (MF,INF).
This is the end of the preview. Sign up
to
access the rest of the document.
 Fall '08
 JOSHUA
 Math

Click to edit the document details